Laplace transform of a piecewise function: Solved exercise

Proceeding in the same way we did in the previous example:

1. Rewrite the piecewise function as a sum of Heaviside functions

we have:

f_{0}(t)=0,\quad f_{1}(t)=\cos\left(t+\frac{\pi}{3}\right)

and

\displaystyle t_{1}=\frac{\pi}{3}

Hence, f may be rewritten using the unit step function:

\begin{aligned} & f(t)=f_{0}(t)+u(t-t_{1})(f_{1}(t)-f_{0}(t)) \\ & f(t)=0+u\left(t-\frac{\pi}{3}\right)\left(\cos\left(t+\frac{\pi}{3}\right)-0\right)\end{aligned}

which yields:

f(t)=u\left(t-\frac{\pi}{3}\right)\cos\left(t+\frac{\pi}{3}\right)

2. Apply the Laplace transform

We have:

\mathscr{L}\{f(t)\}=\mathscr{L}\left\{u\left(t-\frac{\pi}{3}\right)\cos\left(t+\frac{\pi}{3}\right) \right\}

Before we evaluate the Laplace transform above, note that, from line 10 of the table:

\displaystyle\mathscr{L}\{g(t-a)u(t-a)\}=e^{-as}G(s)

and comparing with the expression of f, we get:

\displaystyle a=\frac{\pi}{3}

and the cosine function can be rewritten as follows

\cos\left(t+\frac{\pi}{3}\right)=\cos\left(t+\frac{2\pi}{3}-\frac{\pi}{3}\right)

which yields:

g(t)=\cos\left(t+\frac{2\pi}{3}\right)

3. Laplace transform of cos (t+2pi/3)

We have:

G(s)=\mathscr{L}\{g(t)\}=\mathscr{L}\left\{\cos\left(t+\frac{2\pi}{3}\right)\right\}

Using the formula:

\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)

we are able to rewrite g(t) as follows:

\begin{aligned} \cos\left(t+\frac{2\pi}{3}\right) & =\cos(t)\cos\left(\frac{2\pi}{3}\right)-\sin(t)\sin\left(\frac{2\pi}{3}\right) \\ & =-\frac{1}{2}\cos(t)-\frac{\sqrt{3}}{2}\sin(t) \end{aligned}

Thus:

G(s)=\mathscr{L}\left\{\cos\left(t+\frac{2\pi}{3}\right)\right\}=-\frac{1}{2}\cdot\mathscr{L}\{\cos(t)\}-\frac{\sqrt{3}}{2}\cdot\mathscr{L}\{\sin(t)\}

For b=1 in lines 3 and 4 of the transform table, we have then:

\begin{aligned} G(s) & =-\frac{1}{2}\cdot\frac{s}{s^{2}+1^{2}}-\frac{\sqrt{3}}{2}\cdot\frac{1}{s^{2}+1^{2}} \\ & =-\frac{1}{2}\cdot\left(\frac{s+\sqrt{3}}{s^{2}+1}\right) \end{aligned}

4. Result

Therefore,

\begin{aligned}\mathscr{L}\{f(t)\} &=e^{-\frac{\pi}{3}s}G(s) \\ \mathscr{L}\{f(t)\} &=-\frac{e^{-\frac{\pi}{3}s}}{2}\left(\frac{s+\sqrt{3}}{s^{2}+1}\right) \end{aligned}

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